The Area Is The Same

We can use the disc method to find the volume of the given solids of revolution.

$V = \int_a^b dV = \int_a^b \pi (f(x))^2 \, dx$

To find the volume of solid A, we can find the volume of revolution of $y = \frac{1}{2} x$ about the $x$ -axis in the interval $0 \leq x \leq a$ .

$V_{\text{A}} = \int _{0} ^{a} \pi \left( \frac{1}{2} x\right)^{2} \, dx$

It evaluates to $\dfrac{a^{3}}{12} \pi$ .

To find the volume of solid B, we can find the volume of revolution of $y = x$ about the $x$ -axis, and then subtract the volume of revolution of $y = \frac{1}{2} x$ , i.e. volume of solid A from it.

$V_{\text{B}} = \int _{0} ^{a} \pi x^{2} \, dx - \int _{0} ^{a} \pi \left( \frac{1}{2} x\right)^{2} \, dx$

It evaluates to $\dfrac{a^{3}}{3}\pi - \dfrac{a^{3}}{12}\pi = \dfrac{a^{3}}{4} \pi$ .

Hence volume of solid B is greater than volume of solid A. $_\square$

When the blue area is revolved it will form cone.Since it satisfy $y=\frac{ x}{2}$ ,when $x=a$ then $y=\frac{a}{2}$ .Hence the cone formed has radius $\frac{a}{2}$ and height $a$ .

Volume of the cone( $formed$ $by$ $blue$ $region$ ) $=$ $\frac{1}{3} \times \pi \times a(\frac{a}{2})^2$

$\boxed {Volume_{blue}=\frac{\pi a^2}{12}}$

$\boxed {Volume_{red}=Volume_{red+blue}-Volume_{blue}}$

$Volume_{red}=\frac{\pi (a^2)a}{3}-\frac{\pi a^3}{12}$

$\boxed {Volume_{red}=\frac{\pi a^3}{3}[\frac{3}{4}]}$

$\boxed{Volume_{red}>Volume_{blue}}$

Consider Solid A:Solid (A+B)
They have same height but A has half the radius.
$A:A+B=1:4$
$A:B=1:3$
$B$ is larger.

very simple: witch y is larger? Well we know y#1 is between 0 and 1/2x, and y#2 is between 1/2x and 1x

Using logic we can figure out that 3/4x is larger than 1/4x

So the answer is B